Question Number 210761 by mr W last updated on 18/Aug/24
Commented by mr W last updated on 18/Aug/24
$${find}\:{the}\:{radius}\:{R}\:{of}\:{the}\:{largest}\:{sphere} \\ $$$${which}\:{can}\:{be}\:{placed}\:{between}\:{the}\:{red} \\ $$$${and}\:{blue}\:{planes}\:{in}\:{the}\:{first}\:{octant}. \\ $$$${assume}\:\mathrm{0}<{p}\leqslant{a},\:\mathrm{0}<{q}\leqslant{b},\:\mathrm{0}<{r}\leqslant{c}. \\ $$
Commented by mahdipoor last updated on 18/Aug/24
$${planes}\:{is}\:{not}\:{infinity}\:? \\ $$
Commented by mr W last updated on 18/Aug/24
$${the}\:{sphere}\:{must}\:{completely}\:{lie}\:{in} \\ $$$${the}\:{first}\:{octant}\:{and}\:{between}\:{the} \\ $$$${two}\:{planes}.\:{i}.{e}.\:{the}\:{sphere}\:{must} \\ $$$${lie}\:{inside}\:{a}\:{region}\:{bounded}\:{by}\:\mathrm{5} \\ $$$${planes}. \\ $$
Commented by mr W last updated on 19/Aug/24
$${the}\:{question}\:{is}\:{equivalent}\:{to}\:{following} \\ $$$${question}:\:{find}\:{the}\:{radius}\:{of}\:{the} \\ $$$${largest}\:{sphere}\:{which}\:{can}\:{be}\:{cut} \\ $$$${from}\:{following}\:{solid}\:{object}: \\ $$
Commented by mr W last updated on 19/Aug/24
Commented by ajfour last updated on 25/Aug/24
Commented by mr W last updated on 27/Aug/24
$${please}\:{check}\:{my}\:{approach}\:{in}\:{Q}\mathrm{211058} \\ $$
Answered by BHOOPENDRA last updated on 18/Aug/24
$${R}={min}\left(\frac{{p}}{\mathrm{2}}\:,\frac{{q}}{\mathrm{2}},\frac{{r}}{\mathrm{2}}\right) \\ $$$${This}\:{would}\:{be}\:{the}\:{the}\:{radius}\:{that} \\ $$$${can}\:{be}\:{placed}\:{between}\:{the}\:{two}\:{planes} \\ $$$${i}\:{got}\:{this}\:{Mr}.{W}\:\:. \\ $$$${i}\:{wrote}\:{the}\:{eqution}\:{of}\:{the}\:{red}\:\&{blue} \\ $$$${planes}\:{for}\:{the}\:{positioning}\:{of}\:{sphere} \\ $$$${the}\:{largest}\:{sphere}\:{that}\:{can}\:{fit}\:{between} \\ $$$${these}\:{planes}\:{is}\:{equidistance}\:{from}\: \\ $$$${both}\:{planes}.{The}\:{radius}\:{R}\:{of}\:{this}\: \\ $$$${sphere}\:{will}\:{be}\:{equal}\:{to}\:{shortest}\:{distnce} \\ $$$${between}\:{two}\:{planes}\:{divided}\:{by}\:\mathrm{2}. \\ $$$${If}\:{this}\:{is}\:{correct}\:{Ans}\:{then}\:{i}\:{can}\: \\ $$$${do}\:{further}\:{calculations}. \\ $$
Commented by mr W last updated on 19/Aug/24
$${i}\:{don}'{t}\:{have}\:{the}\:{solution}.\:{if}\:{you} \\ $$$${think}\:{you}\:{have}\:{a}\:{solution},\:{just} \\ $$$${post}\:{it}.\:{thanks}! \\ $$
Commented by BHOOPENDRA last updated on 19/Aug/24
Answered by BHOOPENDRA last updated on 19/Aug/24
$${Red}\:{plane} \\ $$$$\frac{{x}}{{p}}+\frac{{y}}{{q}}+\frac{{z}}{{r}}=\mathrm{1} \\ $$$${Normal}\:{vector}\:\overset{\rightarrow} {{n}}_{\mathrm{1}} =\left(\frac{\mathrm{1}}{{p}},\frac{\mathrm{1}}{{q}},\frac{\mathrm{1}}{{r}}\right){Or}\: \\ $$$$\left({A}_{\mathrm{1}} ,{B}_{\mathrm{1}} ,{C}_{\mathrm{1}} ,{D}_{\mathrm{1}} \right) \\ $$$${for}\:{the}\:{blue}\:{plane} \\ $$$$\frac{{x}}{{a}}+\frac{{y}}{{b}}+\frac{{z}}{{c}}=\mathrm{1} \\ $$$${Normal}\:{vector}\:\overset{\rightarrow} {{n}}_{\mathrm{2}} =\left(\frac{\mathrm{1}}{{a}},\frac{\mathrm{1}}{{b}},\frac{\mathrm{1}}{{c}}\right)\:{Or} \\ $$$$\left({A}_{\mathrm{2}} ,{B}_{\mathrm{2}} ,{C}_{\mathrm{2}} ,{D}_{\mathrm{2}} \right) \\ $$$${the}\:{distnace}\:{between}\:{two}\:{planes} \\ $$$${Ax}+{By}+{Cz}={D}_{\mathrm{1}} \:\&{Ax}+{By}+{Cz}={D}_{\mathrm{2}} \\ $$$$\:{d}=\frac{\mid{D}_{\mathrm{1}} −{D}_{\mathrm{2}} \mid}{\:\sqrt{{A}^{\mathrm{2}} +{B}^{\mathrm{2}} +{C}^{\mathrm{2}} }} \\ $$$${R}={minmum}\:{distnace}\:{between}\:{planes}\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${If}\:{the}\:{planes}\:{are}\:{far}\:{from}\:{each}\:{other} \\ $$$${Second} \\ $$$${R}={min}\left(\frac{{p}}{\mathrm{2}},\frac{{q}}{\mathrm{2}},\frac{{r}}{\mathrm{2}}\right)\:{or}\left(\frac{{a}}{\mathrm{2}},\frac{{b}}{\mathrm{2}},\frac{{c}}{\mathrm{2}}\right) \\ $$$${Assuming}\:{the}\:{sphere}\:{is}\:{centered} \\ $$$${at}\:{midpoint}\:{between}\:{the}\:{planes}.{And} \\ $$$${planes}\:{are}\:{parellel}\:{to}\:{each}\:{other}. \\ $$$$ \\ $$$$ \\ $$
Commented by mr W last updated on 19/Aug/24
$${only}\:{when}\:{two}\:{planes}\:{are}\:{parallel},\: \\ $$$${we}\:{can}\:{speak}\:{of}\:{the}\:“{distance}''\: \\ $$$${between}\:{two}\:{planes}.\:{in}\:{general} \\ $$$${case},\:{the}\:{two}\:{planes}\:{are}\:{not}\:{parallel} \\ $$$${to}\:{each}\:{other},\:{and}\:{we}\:{can}\:{place}\:{a} \\ $$$${infinitely}\:{large}\:{sphere}\:{between} \\ $$$${them}.\:{but}\:{since}\:{the}\:{sphere}\:{must} \\ $$$${lie}\:{in}\:{the}\:{first}\:{octant},\:{the}\:{size}\:{of} \\ $$$${the}\:{sphere}\:{is}\:{limited}. \\ $$$${besides}\:{we}\:{can}\:{see},\:{when}\:{p},{q},{r}\:{are} \\ $$$${very}\:{small},\:{the}\:{radius}\:{of}\:{the}\:{sphere} \\ $$$${is}\:{only}\:{dependent}\:{from}\:{a},{b},{c}. \\ $$$${your}\:{solution}\:{seems}\:{not}\:{to}\:{be}\: \\ $$$${correct}. \\ $$
Commented by BHOOPENDRA last updated on 19/Aug/24
$${You}\:{are}\:{right}\:{Mr}.{W}\:\:{i}\:{considered}\: \\ $$$${planes}\:{are}\:{parallel}. \\ $$